Study Notes on Newton's Second Law and Atwood Machine Application

Introduction to Newton's Second Law

• Application of Newton's Second Law to systems of objects, such as the Atwood machine.

• Instructor: Kristin Gonzalez Vega, Centennial High School, Frisco, Texas.

Example Scenario

• Two objects (blocks) with masses denoted as M1 and M2 connected by a string which is pulled by an external force, F0, to the right.

Problem Setup

• Objective: Find the acceleration.

• Approach: Draw free body diagrams for each mass individually.

Free Body Diagrams

Mass 1 (M1)

• Forces acting on M1:

  • Normal force (N1) acting upwards.

  • Gravitational force (M1 * g) acting downwards.

• Equation can be established:

  • In the Y direction: N1 = M1 * g (no vertical motion, hence the forces are balanced).

  • In the X direction:

  • Tension (T) acting to the right = M1 * a.

Mass 2 (M2)

• Forces acting on M2:

  • Pull force (F0) acting to the right.

  • Tension (T) acting to the left.

• Equation can be established:

  • In the X direction: F0 - T = M2 * a.

  • In the Y direction:

  • Normal force (N2) balances M2 * g downwards (forces balanced).

System Approach vs Individual Approach

• Tension Calculation: Both equations have Tension (T) and acceleration (a) as unknowns, leading to complexity.

• Alternatively, treat the two objects as a single system.

• Center of Mass: Draw forces acting on the combined mass:

  • Normal force acts upwards.

  • Combined gravitational force down: (M1 + M2) * g.

  • F0 is the only external force acting in the horizontal direction.

Simplification through System Approach

• For the system:

  • Y direction: N = (M1 + M2) * g (balances out).

  • X direction: F0 = (M1 + M2) * a.

• Final equation for the system:

  • F0 = (M1 + M2) * a, where total mass is M1 + M2.

• Tension becomes an internal force and cancels out when analyzing the system as one entity.

Atwood Machine Application

• Setup: Atwood machine consists of a pulley system with two masses.

• One known mass and an unknown mass to be determined using acceleration measurements.

Analyzing Motion in the Atwood Machine

• Release system from rest; it will accelerate.

• Connect the force interactions through free body diagrams for each mass:

  • Left mass (M1): Down: M1 * g, Up: Tension (T).

  • Equation if moving down: M1 * g - T = M1 * a.

  • Right mass (M2): Upwards: Tension (T), Down: M2 * g.

  • Equation: T - M2 * g = M2 * a.

• Since tension is identical in a massless pulley system, consider behavior as a system for simplification.

Finding System Acceleration

• External forces acting on the system:

  • M1 * g downward and M2 * g upward.

• Total force: M1 * g - M2 * g = (M1 + M2) * a.

• Solve for acceleration (a):

  • Use graphing techniques (velocity vs. time graph) to derive acceleration from slope.

• Example of measured acceleration: a = 1.22 m/s².

Calculation of Unknown Mass

• Substitute back into derived equations to solve:

• Known mass (M2) = 0.05 kg.

• Use: (M1 * g) - (M2 * g) = (M1 + M2) * 1.22 m/s².

• After calculation: M1 ≈ 0.064 kg.

• Comparison with real measured value: 0.065 kg.

Finding Tension in the String

• After establishing acceleration and known masses, backtrack to find tension if needed:

  • Using the equation for the larger mass:

  • M1 * g - T = M1 * a.

• Tension calculated ensuring all variables accounted.

Summary and Key Takeaways

• System Approach: More efficient for solving acceleration.

• Individual Force Analysis: Necessary for specific force calculations like tension.

• Correct Methodology: Always write net force equations derived from free body diagrams to avoid skipping steps in calculations.